How are these equations of SGD with momentum equivalent?
The first two equations are equivalent. The last equation can be equivalent if you scale $\alpha$ appropriately.
Equation 1
Consider the equation from the Stanford slide:
$$
v_{t}=\rho v_{t-1}+\nabla f(x_{t-1}) 
\\ 
x_{t}=x_{t-1}-\alpha v_{t},
$$
Let's evaluate the first few $v_t$ so that we can arrive at a closed form solution:
$v_0 = 0 \\
v_1 =  \rho v_0 + \nabla f(x_0) = \nabla f(x_0)\\
v_2 = \rho v_1 + \nabla f(x_1) = \rho \nabla f(x_0) + \nabla f(x_1)\\
v_3 = \rho v_2 + \nabla f(x_2) = \rho^2 \nabla f(x_0) + \rho \nabla f(x_1) + \nabla f(x_2)\\
\dots \\
v_t = \displaystyle \sum_{i=0}^{t-1} \rho^{t-1-i} \nabla f(x_i)
$
So the closed form update is:
$$x_t = x_{t-1} - \alpha \displaystyle \sum_{i=0}^{t-1} \rho^{t-1-i} \nabla f(x_i)$$
Equation 2
Now consider the equation from the paper:
$$
v_{t}=\rho v_{t-1}+\alpha \nabla f(x_{t-1}) 
\\ 
x_{t}=x_{t-1}- v_{t},
$$
We again evaluate the first few $v_t$ to arrive at a closed form solution:
$v_0 = 0 \\
v_1 =  \rho v_0 + \alpha \nabla f(x_0) = \alpha \nabla f(x_0)\\
v_2 = \rho v_1 + \alpha \nabla f(x_1) = \rho \alpha \nabla f(x_0) + \alpha \nabla f(x_1)\\
v_3 = \rho v_2 + \alpha \nabla f(x_2) = \rho^2 \alpha \nabla f(x_0) + \rho \alpha \nabla f(x_1) + \alpha \nabla f(x_2)\\
\dots \\
v_t = \displaystyle \sum_{i=0}^{t-1} \rho^{t-1-i} \alpha \nabla f(x_i)
$
So the closed from update is:
$$x_t = x_{t-1} - \displaystyle \sum_{i=0}^{t-1} \rho^{t-1-i} \alpha \nabla f(x_i)$$
As you can see, this is equivalent to the previous closed form update. The only difference is if $\alpha$ is inside or outside the summation, but since it is a constant, it doesn't really matter anyways.
Equation 3
As for the last equation
$$
v_{t}= \rho v_{t-1}+ (1- \rho) \nabla f(x_{t-1}) 
\\ 
x_{t}=x_{t-1}-\alpha v_{t}
$$
Let's do the same thing:
$v_0 = 0 \\
v_1 =  \rho v_0 + (1-\rho) \nabla f(x_0) = (1-\rho) \nabla f(x_0)\\
v_2 = \rho v_1 + (1-\rho) \nabla f(x_1) = \rho (1-\rho) \nabla f(x_0) + (1-\rho) \nabla f(x_1)\\
v_3 = \rho v_2 + (1-\rho) \nabla f(x_2) = \rho^2 (1-\rho) \nabla f(x_0) + \rho (1-\rho) \nabla f(x_1) + (1-\rho) \nabla f(x_2)\\
\dots \\
v_t = \displaystyle \sum_{i=0}^{t-1} \rho^{t-1-i} (1-\rho) \nabla f(x_i)
$
And so the closed form update is:
$$x_t = x_{t-1} - \alpha \displaystyle \sum_{i=0}^{t-1} \rho^{t-1-i} (1-\rho) \nabla f(x_i)$$
This equation is equivalent to the other two as long as you scale $\alpha$ by a factor of $\displaystyle \frac{1}{1-\rho}$.